find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$  so $z=\frac{-11}{8}$  however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you
 A: There is a simpler method. Just use
$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$
To find the complex roots, just  solve the quadratic:
$$a^2+ab+b^2=0.$$

In your case you can take $$a=2z+3, b=\dfrac 14.$$
A: $$(2z+3)^3=\frac{1}{64} \implies 2z+3=\frac{1}{4}, \frac{\omega}{4}, \frac{\omega^2}{4}$$
$$\implies z=\frac{-11}{8}, \frac{\omega -12}{8}, \frac{\omega^2-12}{8}.$$ Here $\omega$ is cube-root of unity.
A: For any $z = a + ib \in \mathbb{C}$,
$$
z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right),
$$
where
$$
\rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right).
$$
Then,
$$
\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\right)}\right), \quad k = 0, 1, \ldots, n-1.
$$
In our example,
$$
\frac{1}{64} = \frac{1}{64} + i0, \text{ i. e., }a = \frac{1}{64}\text{ and }b = 0 \Rightarrow
$$
$$
\frac{1}{64} = \frac{1}{64}\left(\cos(0) + i\sin(0)\right) \Rightarrow
$$
$$
\sqrt[3]{\frac{1}{64}} = \sqrt[3]{\frac{1}{64}}\left(\cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)\right), \quad k = 0, 1, 2.
$$
$$
\begin{array}{rl}
k = 0 \Rightarrow & z_1 = \frac{1}{4} \\
k = 1 \Rightarrow & z_2 = \frac{1}{4}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1 - i\sqrt{3}}{8}\\
k = 2 \Rightarrow & z_3 = \frac{1}{4}\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1 + i\sqrt{3}}{8}
\end{array}
$$
A: Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots
While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get the remaining roots
$$ z^2+z+1$$
$$ z_1 = -\frac{1}{2}  + \frac{\sqrt{ -3}}{2}$$
$$ z_2 = -\frac{1}{2}  - \frac{\sqrt{ -3}}{2}$$
They are conjugate of eachother, If you cube it using the property of complex number the result is $z^3 = 1$, so $z_0 , z_1, z_2$ all from the cube root of unity
Anytime you take the cube root of a number $n$, the solutions are $\sqrt[3]{n} , \sqrt[3]{n} \cdot z_1, \sqrt[3]{n} \cdot z_2$
